$H$ is a Hilbert space, and $T\in B(H)$ continuous and Fredholm operator.
Same books in definition of Fredholm use (when work in Banach space)
$\operatorname{ind}(T)=\dim(\ker(T))-\dim(\operatorname{coker}(T))$
and another's (when work with Hilbert space)
$\operatorname{ind}(T)=\dim(\ker(T))-\dim(\ker(T^{*}))$ , $T^{*}$ is the adjoint operator
so how can i show that
$\dim(\operatorname{coker}(T))=\dim(\ker(T^{*}))$
$\operatorname{coker}(T)=H/T(H)$ and $Ker(T^{*})=(T(H))^{\perp}$. so in Hilbert is true $H/T(H)\approx (T(H))^{\perp}$ ?
thanks
Best Answer
Proof.
$\newcommand{\co}{\operatorname{coker}}$Let \begin{align*}\Phi: & K^\perp\to H/K\\ & h\mapsto [h]\end{align*}
(1) Of course, $\Phi$ is linear bounded.
(2) $\Phi$ is injective. Suppose $h\in K^\perp$ such that $\Phi(h)=0$, then $h\in K^\perp\cap K$, so $h=0$.
(3) $\Phi$ is surjective. For any $h\in H$, there is some $h'\in K^\perp$ such that $h-h'\in K$, thus $\Phi(h')=[h']=[h]$.
Noting that $TH$ is closed subspace in the question, $H/TH\cong (TH)^\perp$.